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  "body": "\n## Statement\n\nLet $G$ be a finite [[planar graph|029JQEK15B]] that contains no triangle, that is, no cycle of length $3$.\nThen $G$ is [[$3$-colourable|0Y7VDAX8Q3]].\n\n## Remarks\n\nAn easy weaker bound comes from [[Euler's formula|0W1C598HMS]].\nA triangle-free planar graph with $n \\ge 3$ vertices has at most $2n - 4$ edges, so it has a vertex of degree at most $3$.\nInduction then gives a colouring with $4$ colours.\nGetting down to $3$ colours needs a finer induction, which proves a stronger statement about extending colourings from a short precoloured face.\nHerbert Grötzsch proved the theorem in 1959, and Carsten Thomassen later gave shorter proofs.\nBoth hypotheses are needed.\nWithout the triangle condition the bound fails: $K_4$ is planar and needs $4$ colours.\nWithout planarity it fails too: the Grötzsch graph, with $11$ vertices, is triangle-free and needs $4$ colours.\nFor example, every planar bipartite graph is triangle-free, and indeed $2$-colourable.\n\n## Explanations\n\n### L1 human\n\nThe four colour theorem says that $4$ colours suffice for every planar graph.\nIf the graph has no triangles, Grötzsch's theorem says that $3$ colours suffice.\nTriangles are the main reason a planar graph might need a fourth colour.\n\n### L2 human\n\nThe theorem is a sharp statement about the interplay of girth and colouring on surfaces.\nGrünbaum and Aksenov extended it to planar graphs with at most three triangles.\nThe precoloured-face strengthening used in the proof is typical of colouring proofs for planar graphs, as in the proof that planar graphs are $5$-choosable.\n",
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